PhysicsISC Class 12 · Moving Charges and Magnetism

Motion of Charged Particles in Magnetic and Electric Fields

Updated 15 October 2026 · Fact-checked

A charge q moving with velocity $\vec{v}$ through fields $\vec{E}$ and $\vec{B}$ feels the Lorentz force $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$. The magnetic part, perpendicular to $\vec{v}$, does no work. Moving perpendicular to a uniform B, the charge follows a circle of radius $r = \frac{mv}{qB}$ with frequency $f = \frac{qB}{2\pi m}$, independent of its speed.

Force is always at right angles to velocity, so it points to the centre, changes only direction, does no work, and gives r = mv/qB.

Illustration drawn for QuestionBuddy.

Understand Motion of Charges in a Magnetic Field

Magnetic force on a moving charge. $F_{B} = q (v \times B)$ , with magnitude $F = q v B sin θ$ . The order matters: $v \times B$ , not $B \times v$ (the 2024 ISC 1-mark item tested exactly this). The force is:

zero if the charge is at rest or moves along the field (θ = 0° or 180°);
maximum, $q v B$ , when $v ⊥ B$ ;
always perpendicular to both $v$ and $B$ , so it does no work and cannot change the speed or kinetic energy, only the direction.

For a negative charge the force is opposite to that on a positive charge. Unit check: 1 T = 1 N s C⁻¹ m⁻¹ = 1 N A⁻¹ m⁻¹.

Lorentz force. With both fields present, $F = q E + q (v \times B)$ .

**Circular motion ( $v ⊥ B$ ).** The magnetic force supplies the centripetal force:
$q v B = \frac{m v ^{2}}{r} \Rightarrow r = \frac{m v}{q B} = \frac{p}{q B} .$
Period and frequency:
$T = \frac{2 π r}{v} = \frac{2 π m}{q B}, f = \frac{q B}{2 π m}, ω = \frac{q B}{m} .$
The period does not depend on speed or radius: faster particles move in bigger circles but take the same time. In terms of kinetic energy, $r = \frac{2 m K}{q B}$ , and for a particle accelerated from rest through p.d. V, $K = q V$ .

Sense of rotation. For B along −z and a particle moving along +x (as in the 2025 question), the force on a positive charge is $q v B (\hat{i} \times - \hat{k}) = + q v B \hat{j}$ , so a proton curves towards +y: it circles in the xy-plane. An electron curves the opposite way. The path is a circle, not a helix, because there is no velocity component along B. (The 2025 examiners noted answers of "helical" or "linear".) A positive charge circles clockwise when viewed from the side the field points towards.

Helical motion. If $v$ makes angle θ with $B$ , split it: $v_{⊥} = v sin θ$ produces circular motion of radius $r = \frac{m v s i n θ}{q B}$ ; $v_{∥} = v cos θ$ is unaffected. The result is a helix with
$pitch = v_{∥} T = \frac{2 π m v c o s θ}{q B} .$

Crossed fields: velocity selector. With $E ⊥ B$ , both perpendicular to $v$ , the electric force qE and magnetic force qvB can be arranged to oppose. Particles pass undeflected when $q E = q v B$ , that is $v = \frac{E}{B}$ , whatever their charge or mass.

Force on a current-carrying conductor. A wire of length l carrying current I contains moving charges. Replacing $q v$ by $I d l$ (since $q v = q \frac{d l}{d t} = I d l$ ):
$d F = I d l \times B, F = I l \times B, F = B I l sin θ .$
Direction by the right-hand rule for the cross product (the syllabus asks for right-hand rules only). A straight wire parallel to B feels no force.

Motion of Charges in a Magnetic Field: formulas and key facts

Lorentz force: $F = q (E + v \times B)$; Order: v × B.
Magnetic force: $F = q v B sin θ$; Does no work.
Radius: $r = \frac{m v}{q B} = \frac{2 m K}{q B}$; $v ⊥ B$ .
Period and frequency: $T = \frac{2 π m}{q B}, f = \frac{q B}{2 π m}$; Independent of speed.
Helix pitch: $p = \frac{2 π m v c o s θ}{q B}$; Radius uses $v sin θ$ .
Velocity selector: $v = \frac{E}{B}$; Crossed fields, no deflection.
Force on a wire: $F = I l \times B$; $F = B I l sin θ$ .

How to solve Motion of Charges in a Magnetic Field questions

Split every problem into "magnitude" and "direction".

1Write $v$ and $B$ as vectors (or note the angle between them).
2Find the magnitude $q v B sin θ$ ; for a wire, $B I l sin θ$ .
3Find the direction from $v \times B$ (unit-vector cross products), reversing it for negative charges.
4For circular motion, equate $q v B$ to $\frac{m v ^{2}}{r}$ ; get r, then T or f.
5If $v$ is oblique to $B$ , use $v sin θ$ for the radius and $v cos θ$ for the pitch.
6For crossed fields, balance qE against qvB.

Motion of Charges in a Magnetic Field in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Radius of a charged particle's path

Suppose an electron of kinetic energy 100 eV moves perpendicular to a weak magnetic field of $5 \times 1 0^{- 5}$ T in a large vacuum chamber.

Find its speed and the radius of its circular path.

Try it first, then show the solutionSolution

$K = 100 \times 1.6 \times 1 0^{- 19} = 1.6 \times 1 0^{- 17}$ J; $v = \frac{2 K}{m} = \frac{3.2 \times 1 0 ^{- 17}}{9.1 \times 1 0 ^{- 31}} \approx 5.93 \times 1 0^{6}$ m s⁻¹.
$r = \frac{m v}{e B} = \frac{9.1 \times 1 0 ^{- 31} \times 5.93 \times 1 0 ^{6}}{1.6 \times 1 0 ^{- 19} \times 5 \times 1 0 ^{- 5}} \approx 0.67$ m.

Answer: About $5.9 \times 1 0^{6}$ m s⁻¹ and a radius of about 0.67 m.

Memory hook: r = momentum ÷ (charge × field).

Deflecting an electron beam

Imagine an old cathode-ray display in which electrons travel at $2 \times 1 0^{7}$ m s⁻¹ through a deflecting field of $1 \times 1 0^{- 3}$ T at right angles to the beam.

Find the force on each electron and its acceleration.

Try it first, then show the solutionSolution

$F = e v B = 1.6 \times 1 0^{- 19} \times 2 \times 1 0^{7} \times 1 0^{- 3} = 3.2 \times 1 0^{- 15}$ N.
$a = \frac{F}{m} = \frac{3.2 \times 1 0 ^{- 15}}{9.1 \times 1 0 ^{- 31}} \approx 3.5 \times 1 0^{15}$ m s⁻².

Answer: $3.2 \times 1 0^{- 15}$ N; about $3.5 \times 1 0^{15}$ m s⁻², always sideways to the motion.

Memory hook: Magnetic force turns, never speeds up.

Tricks to remember Motion of Charges in a Magnetic Field

Short memory aids. Say each one aloud once, then use it on a question the same day.

v cross B, in that order
"Velocity first, field second" for the force on a charge; "length first, field second" for a wire.
Use it when: Vector form of the force.
Same time, bigger circle
Faster particles circle wider but in the same time: $T = \frac{2 π m}{q B}$ .
Use it when: Frequency questions.
No work, no speed change
Force ⊥ velocity always, so KE is constant.
Use it when: Assertion-reason items.
E over B
Undeflected speed in crossed fields = E ÷ B.
Use it when: Velocity selector.

Motion of Charges in a Magnetic Field: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: The Lorentz force in vector form · 1 mark · MCQBoard level

The Lorentz force on a charge q moving with velocity $v$ in fields $E$ and $B$ is:
(a) $q (E + B \times v)$
(b) $q (E + v \times B)$
(c) $q (E \cdot v + B)$
(d) $q E \times v B$

Try it first, then show the solutionSolution

The magnetic part is $q v \times B$ ; reversing the order reverses the force.

Answer: (b).

Example 2: Radius and frequency of an electron · 2 marks · Short answerBoard level

An electron moves at $3 \times 1 0^{6}$ m s⁻¹ perpendicular to a uniform field of $2 \times 1 0^{- 3}$ T. Find the radius of its path and its frequency of revolution. ( $m_{e} = 9.1 \times 1 0^{- 31}$ kg, $e = 1.6 \times 1 0^{- 19}$ C)

Try it first, then show the solutionSolution

$r = \frac{m v}{e B} = \frac{9.1 \times 1 0 ^{- 31} \times 3 \times 1 0 ^{6}}{1.6 \times 1 0 ^{- 19} \times 2 \times 1 0 ^{- 3}} \approx 8.5 \times 1 0^{- 3}$ m. (1)
$f = \frac{e B}{2 π m} = \frac{3.2 \times 1 0 ^{- 22}}{2 π \times 9.1 \times 1 0 ^{- 31}} \approx 5.6 \times 1 0^{7}$ Hz. (1)

Answer: About 8.5 mm; about 56 MHz.

Example 3: Direction of force and sense of rotation · 3 marks · Short answerHarder board level

A proton moves along +y in a uniform field along +z. Find the direction of the force on it, describe its path and the sense of rotation seen from +z. How would an electron with the same velocity move?

Try it first, then show the solutionSolution

$F = e (v \hat{j} \times B \hat{k}) = e v B \hat{i}$ : along +x. (1)
The force is always perpendicular to the velocity in the xy-plane, so the proton moves in a circle in that plane. Starting along +y and turning towards +x, it goes clockwise as seen from +z. (1)
The electron's force is along −x: it circles in the xy-plane anticlockwise (seen from +z), in a smaller circle because its mass is smaller. (1)

Answer: Force along +x; proton circles clockwise, electron anticlockwise (seen from +z), both in the xy-plane.

Example 4: A helical path · 3 marks · Short answerHarder board level

A proton with speed $4 \times 1 0^{5}$ m s⁻¹ enters a uniform field of 0.1 T at 60° to the field. Find the radius of the helix, the period and the pitch. ( $m_{p} = 1.67 \times 1 0^{- 27}$ kg)

Try it first, then show the solutionSolution

$v_{⊥} = v sin 60° \approx 3.46 \times 1 0^{5}$ m s⁻¹; $v_{∥} = v cos 60° = 2 \times 1 0^{5}$ m s⁻¹. (1)
$r = \frac{m v _{⊥}}{e B} = \frac{1.67 \times 1 0 ^{- 27} \times 3.46 \times 1 0 ^{5}}{1.6 \times 1 0 ^{- 20}} \approx 3.6$ cm; $T = \frac{2 π m}{e B} \approx 6.56 \times 1 0^{- 7}$ s. (1)
Pitch $= v_{∥} T \approx 2 \times 1 0^{5} \times 6.56 \times 1 0^{- 7} \approx 0.13$ m. (1)

Answer: r ≈ 3.6 cm, T ≈ 0.66 µs, pitch ≈ 13 cm.

Example 5: Floating a rod with a magnetic force · 3 marks · Short answerHOTS / competency

A horizontal metal rod of length 20 cm and mass 30 g hangs from two light flexible wires in a horizontal field of 0.5 T perpendicular to the rod. What current, and in which sense, will make the tension in the wires zero? (g = 9.8 m s⁻²)

Try it first, then show the solutionSolution

For zero tension, the magnetic force must be upward and equal to the weight: $B I l = m g$ . (1)
$I = \frac{m g}{B l} = \frac{0.03 \times 9.8}{0.5 \times 0.2} = 2.94$ A. (1)
Direction: $F = I l \times B$ must point up; if B points north, the current must flow east (since $\overset{e}{^} \times \overset{n}{^} =$ up). (1)

Answer: About 2.9 A, in the sense that makes $I l \times B$ point upward.

Example 6: Selecting and separating ions · 4 marks · Case-basedHOTS / competency

Suppose singly charged neon ions of mass 20 u and 22 u ( $1 u = 1.66 \times 1 0^{- 27}$ kg) pass undeflected through crossed fields E = $1.5 \times 1 0^{4}$ V m⁻¹ and B = 0.05 T, then enter a region of field 0.1 T where they move in semicircles.
(i) Find the speed of the ions that get through. (1 mark)
(ii) Find the radius of each ion's path. (2 marks)
(iii) How far apart do they land after a semicircle? (1 mark)

Try it first, then show the solutionSolution

(i) $v = \frac{E}{B} = \frac{1.5 \times 1 0 ^{4}}{0.05} = 3 \times 1 0^{5}$ m s⁻¹.
(ii) $r_{20} = \frac{20 \times 1.66 \times 1 0 ^{- 27} \times 3 \times 1 0 ^{5}}{1.6 \times 1 0 ^{- 19} \times 0.1} \approx 0.62$ m.
$r_{22} = \frac{22}{20} \times 0.62 \approx 0.68$ m.
(iii) Separation $= 2 (r_{22} - r_{20}) \approx 0.12$ m.

Answer: (i) $3 \times 1 0^{5}$ m s⁻¹; (ii) about 0.62 m and 0.68 m; (iii) about 12 cm.

Where students lose marks in Motion of Charges in a Magnetic Field

The slips examiners see most often, and what to write instead.

Mistake: Wrong order in the cross product
The ISC 2024 examiners noted candidates choosing $B \times v$ through not understanding vector order.
Do this: Learn $F = q (v \times B)$ and $F = I l \times B$ exactly.
Usually costs: 1 mark.
Mistake: Helical or straight path when v ⊥ B
In 2025 candidates wrote helical or linear for a particle moving perpendicular to B.
Do this: Helix only when there is a velocity component along B; perpendicular gives a circle.
Usually costs: 1 mark.
Mistake: Not knowing f = qB/2πm; units and rounding
Also in 2025: formula unknown, unit omitted, values rounded twice.
Do this: Derive f from $q v B = \frac{m v ^{2}}{r}$ , keep full precision until the end and write Hz.
Usually costs: 1 to 2 marks.
Mistake: Forgetting the sign of the charge
The force on an electron is opposite to $v \times B$ .
Do this: Find $v \times B$ , then reverse it for negative charges.
Usually costs: 1 mark.

Exam tips

Give the sense of rotation (clockwise or anticlockwise, and from which side) when asked to describe the path.
Use unit-vector cross products for direction: $\hat{i} \times \hat{j} = \hat{k}$ and cyclic.
State that kinetic energy stays constant whenever a question involves only a magnetic field.
Cyclotron is not in the 2027 syllabus; frequency questions are about simple circular motion.

Practice questions on Motion of Charges in a Magnetic Field

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The work done by the magnetic force on a charged particle moving in a uniform magnetic field is:
(a) qvB
(b) positive
(c) negative
(d) zero
Hint
Force ⊥ velocity.
Answer
(d).
Q2 · 2 marks · Short answerBoard level
A proton moving along +x enters a uniform field of 0.5 T along −z. Name its path and find its frequency of revolution. ( $m_{p} = 1.67 \times 1 0^{- 27}$ kg) (A variation on ISC 2025.)
Hint
$f = \frac{q B}{2 π m}$ .
Answer
A circle in the xy-plane; $f = \frac{1.6 \times 1 0 ^{- 19} \times 0.5}{2 π \times 1.67 \times 1 0 ^{- 27}} \approx 7.6 \times 1 0^{6}$ Hz.
Q3 · 1 mark · Very short answerHarder board level
An alpha particle and a proton move with the same speed perpendicular to the same field. Find the ratio of the radii of their paths.
Hint
$r \propto \frac{m}{q}$ .
Answer
$\frac{r _{α}}{r _{p}} = \frac{4 m /2 e}{m / e} = 2$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The period of revolution of a charged particle in a uniform magnetic field does not depend on its speed.
Reason (R): The radius is proportional to the speed, so the time for one circle, $\frac{2 π r}{v}$ , stays the same.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
$T = \frac{2 π m}{q B}$ .
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
A 1.5 m wire carries 6 A along +x in a field of 0.2 T along +y. Find the force on it in vector form.
Hint
$\hat{i} \times \hat{j} = \hat{k}$ .
Answer
$F = 1.8 \hat{k}$ N.
Q6 · 2 marks · Short answerHOTS / competency
An electron moving at $4 \times 1 0^{6}$ m s⁻¹ enters, at right angles, a region 2 cm wide containing a uniform field perpendicular to its velocity. Find the least field that stops it from crossing the region.
Hint
It turns back if r ≤ width.
Answer
$B = \frac{m v}{e d} = \frac{9.1 \times 1 0 ^{- 31} \times 4 \times 1 0 ^{6}}{1.6 \times 1 0 ^{- 19} \times 0.02} \approx 1.14 \times 1 0^{- 3}$ T.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose alpha particles (charge 2e, mass $6.64 \times 1 0^{- 27}$ kg) are accelerated from rest through 2 kV and then enter a uniform field of 0.2 T at right angles.
(i) Find their kinetic energy. (1 mark)
(ii) Find their speed. (1 mark)
(iii) Find the radius and frequency of their circular motion. (1 mark)
(iv) By what factor does the radius change if the accelerating voltage is doubled? (1 mark)
Hint
$K = q V$ .
Answer
(i) $6.4 \times 1 0^{- 16}$ J. (ii) $v = \frac{2 K}{m} \approx 4.39 \times 1 0^{5}$ m s⁻¹. (iii) $r = \frac{m v}{q B} \approx 4.6$ cm; $f \approx 1.53$ MHz. (iv) $r \propto V$ : increases by $2$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Force on a charge moving parallel to B?Show answer
Zero.
2.Radius of circular motion in a magnetic field?Show answer
$r = \frac{m v}{q B}$ .
3.Speed undeflected through crossed fields?Show answer
$v = \frac{E}{B}$ .
4.Force on a 2 m wire carrying 3 A perpendicular to 0.5 T?Show answer
3 N.
5.Does the frequency of revolution depend on speed?Show answer
No: $f = \frac{q B}{2 π m}$ .

Motion of Charges in a Magnetic Field: frequently asked questions

Why can a magnetic field not speed up a charge?

The magnetic force is always perpendicular to the velocity, so it does no work and the kinetic energy cannot change; only the direction of motion changes.

When does a charged particle follow a helix?

When its velocity has components both along and perpendicular to the field. The perpendicular part makes it circle; the parallel part carries it along the field.

How is the force on a wire related to the force on a charge?

A current is moving charge. Replacing q\vec{v} with I\,d\vec{l} in q\vec{v} \times \vec{B} gives d\vec{F} = I\,d\vec{l} \times \vec{B}.

Is the cyclotron in the ISC 2027 syllabus?

No. It was removed in the February 2026 revision; the syllabus keeps the force on moving charges, the Lorentz force and circular motion.

How was this tested recently?

In 2024 a 1-mark item asked for the vector form of the Lorentz force; in 2025 a 3-mark question asked for the path of an electron and the frequency of a proton in a field.